A 25-ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.18 ft/sec, how fast, in ft/sec, is the top of the ladder sliding down the wall, at the instant when the bottom of the ladder is 20 ft from the wall? Answer with 2 decimal places.PLZ HELP

h^2 + x^2 = 625 at t = 0 x=20so h = 15

2 h dh + 2 x dx = 0

dh = -(x/h) dx

dh/dt = -(20/15 ) dx/dt = -(20/15).18

what is the answer? .24?

To find the rate at which the top of the ladder is sliding down the wall, we can use related rates.

Let's denote:
x = distance between the bottom of the ladder and the wall (in ft)
y = height of the ladder on the wall (in ft)

Given that the bottom of the ladder slides away from the wall at a rate of 0.18 ft/sec, we can write:

dx/dt = 0.18 ft/sec (1)

We are asked to find dy/dt, the rate at which the top of the ladder is sliding down the wall when x = 20 ft.

Using the Pythagorean theorem, we have:

x^2 + y^2 = 25^2

Differentiating both sides with respect to time (t), we get:

2x * dx/dt + 2y * dy/dt = 0

At the instant when x = 20 ft (as given), we have:

2(20) * (0.18) + 2y * dy/dt = 0

Simplifying this equation, we can solve for dy/dt:

36 + 2y * dy/dt = 0

2y * dy/dt = -36

dy/dt = -36 / (2y)

Now, we need to find the value of y at x = 20 ft. To do this, we can use the Pythagorean theorem:

20^2 + y^2 = 25^2

400 + y^2 = 625

y^2 = 625 - 400

y^2 = 225

y = 15 ft

Substituting this value into the equation for dy/dt:

dy/dt = -36 / (2 * 15) = -36 / 30 = -1.2 ft/sec (rounded to 2 decimal places)

Therefore, at the instant when the bottom of the ladder is 20 ft from the wall, the top of the ladder is sliding down the wall at a rate of -1.2 ft/sec.

To solve this problem, we can use related rates. Related rates problems involve finding the rate of change of one quantity with respect to another quantity when both quantities are changing.

Let's assign variables:

- Let x represent the distance between the bottom of the ladder and the wall.
- Let y represent the distance between the top of the ladder and the ground.

We are given that dx/dt (the rate at which x is changing) is -0.18 ft/sec (negative since x is decreasing). We need to find dy/dt (the rate at which y is changing) when x = 20 ft.

We are asked to find dy/dt, so we differentiate the equation y^2 = 25^2 - x^2 with respect to time t:

2y * dy/dt = 0 - 2x * dx/dt

Note that dx/dt is given as -0.18 ft/sec and we know y = 25 ft.

Plugging in these values into the equation, we have:

2(25) * dy/dt = 0 - 2(20) * (-0.18)

Simplifying further:

50 * dy/dt = 2(20)(0.18)

50 * dy/dt = 7.2

Now, we can solve for dy/dt:

dy/dt = 7.2/50 = 0.144 ft/sec

So, the top of the ladder is sliding down the wall at a rate of 0.14 ft/sec when the bottom of the ladder is 20 ft from the wall.